具体看 yxc 和 彩铅大佬 的题解，灰常不错！

#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 12, M = 1 << 12, INF = 0x3f3f3f3f;

int n, m;
int d[N][N], g[M];
int f[M][N];

int main()
{
    scanf("%d%d", &n, &m);

    memset(d, 0x3f, sizeof d);
    for (int i = 0; i < n; i ++ ) d[i][i] = 0;

    while (m -- )
    {
        int a, b, w;
        scanf("%d%d%d", &a, &b, &w);
        a -- , b -- ;
        d[a][b] = d[b][a] = min(d[a][b], w);
    }

    // 处理每个联通块(集) 能扩展到的边
    for (int i = 1; i < 1 << n; i ++ ) 
        for (int j = 0; j < n; j ++ )
            if (i >> j & 1)
                for (int k = 0; k < n; k ++ )
                    if (d[j][k] != INF) g[i] |= 1 << k;

    memset(f, 0x3f, sizeof f);
    // 初始化起点花费
    for (int i = 0; i < n; i ++ ) f[1 << i][0] = 0;

    for (int i = 1; i < 1 << n; i ++ ) // 枚举全集
        for (int j = (i - 1) & i; j; j = (j - 1) & i)   // 枚举全集的 非全集子集
            if ((g[j] & i) == i) // i 能够通过 子集 j 延伸而来
            {
                int remain = i ^ j;     // 从子集到全集还需要扩展的节点

                int cost = 0;
                for (int k = 0; k < n; k ++ )
                    if (remain >> k & 1)
                    {
                        int t = INF;
                        for (int u = 0; u < n; u ++ )
                            if (j >> u & 1)
                                t = min(t, d[k][u]); // 选择一条最短的路径
                        cost += t;        
                    }

                for (int k = 1; k < n; k ++ ) f[i][k] = min(f[i][k], f[j][k - 1] + cost * k);

            }

    int res = INF;
    for (int i = 0; i < n; i ++ ) res = min(res, f[(1 << n) - 1][i]);

    printf("%d\n", res);

    return 0;
}